What is the solution for y^'=sin(5x),y(0)=2 ?

The solution for y^'=sin(5x),y(0)=2 is y=-1/5 cos(5x)+11/5

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y^'=sin(5x),y(0)=2

y^{^{'}} = sin (5 x), y (0) = 2

y = - \frac{1}{5} cos (5 x) + \frac{11}{5}

Solution steps

y^{'} (x) = sin (5 x)

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int sin (5 x) d x

\int sin (5 x) d x = - \frac{1}{5} cos (5 x) + c_{1}

y = - \frac{1}{5} cos (5 x) + c_{1}

Apply initial conditions:

y = - \frac{1}{5} cos (5 x) + \frac{11}{5}

y = - \frac{1}{5} cos (5 x) + \frac{11}{5}

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 3 x yz)

(x + 1) \frac{d y}{d x} + (x + 2) y = 8 x e^{- x}

\frac{\partial}{\partial x} (5 x^{3} y z^{2})

\int_{- 1}^{2} x^{3} - 1 d x

What is the solution for y^'=sin(5x),y(0)=2 ?
The solution for y^'=sin(5x),y(0)=2 is y=-1/5 cos(5x)+11/5